Two bodies of mass M and 4M kept at a distance y apart. Where should a small particle of mass m be placed from M so that the net gravitational force on it is zero

Options

A. `(y)/(5)`
B. `(y)/(2)`
C. `(y)/(4)`
D. `(y)/(3)`

Answer

To solve the problem of where to place a small particle of mass \( m \) between two bodies of mass \( M \) and \( 4M \) that are \( y \) distance apart, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup**:
- We have two masses: \( M \) and \( 4M \) separated by a distance \( y \).
- Let the distance from mass \( M \) to the small mass \( m \) be \( x \).
- Consequently, the distance from mass \( 4M \) to the small mass \( m \) will be \( y - x \).

2. **Set Up the Gravitational Forces**:
- The gravitational force exerted by mass \( M \) on mass \( m \) is given by:
\[ F_1 = \frac{GMm}{x^2} \]
- The gravitational force exerted by mass \( 4M \) on mass \( m \) is given by:
\[ F_2 = \frac{G(4M)m}{(y - x)^2} \]

3. **Condition for Equilibrium**:
- For the net gravitational force on mass \( m \) to be zero, these two forces must be equal:
\[ F_1 = F_2 \]
- Thus, we have:
\[ \frac{GMm}{x^2} = \frac{G(4M)m}{(y - x)^2} \]

4. **Cancel Common Terms**:
- We can cancel \( G \) and \( m \) from both sides, leading to:
\[ \frac{M}{x^2} = \frac{4M}{(y - x)^2} \]
- Simplifying this gives:
\[ \frac{1}{x^2} = \frac{4}{(y - x)^2} \]

5. **Cross-Multiply**:
- Cross-multiplying yields:
\[ (y - x)^2 = 4x^2 \]

6. **Expand and Rearrange**:
- Expanding the left side:
\[ y^2 - 2yx + x^2 = 4x^2 \]
- Rearranging gives:
\[ y^2 - 2yx - 3x^2 = 0 \]

7. **Solve the Quadratic Equation**:
- This is a quadratic equation in terms of \( x \):
\[ 3x^2 + 2yx - y^2 = 0 \]
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 2y \), and \( c = -y^2 \).
- The discriminant \( D = (2y)^2 - 4(3)(-y^2) = 4y^2 + 12y^2 = 16y^2 \).
- Thus, \( D = 16y^2 \) implies \( \sqrt{D} = 4y \).

8. **Calculate \( x \)**:
- Plugging into the quadratic formula:
\[ x = \frac{-2y \pm 4y}{6} \]
- This gives two solutions:
\[ x = \frac{2y}{6} = \frac{y}{3} \quad \text{(valid solution)} \]
\[ x = \frac{-6y}{6} = -y \quad \text{(not valid, as distance cannot be negative)} \]

9. **Conclusion**:
- Therefore, the small mass \( m \) should be placed at a distance of \( \frac{y}{3} \) from mass \( M \).

### Final Answer:
The small particle of mass \( m \) should be placed at a distance of \( \frac{y}{3} \) from mass \( M \).